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Difference between revisions of "1976 AHSME Problems/Problem 30"

m (→‎Solution)
m (→‎Solution)
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== Solution ==
 
== Solution ==
 
The first equation suggests the substitution <math>a = x</math>, <math>b = 2y</math>, and <math>c = 4z</math>. Then <math>x = a</math>, <math>y = b/2</math>, and <math>z = c/4</math>. Substituting into the given equations, we get  
 
The first equation suggests the substitution <math>a = x</math>, <math>b = 2y</math>, and <math>c = 4z</math>. Then <math>x = a</math>, <math>y = b/2</math>, and <math>z = c/4</math>. Substituting into the given equations, we get  
−
a + b + c &= 12, \\
+
 
−
\frac{ab}{2} + \frac{bc}{2} + \frac{ac}{2} &= 22, \\
 
−
\frac{abc}{8} = 6,
 
−
\end{align*}
 
−
which simplifies to
 
 
\begin{align*}
 
\begin{align*}
 
a + b + c &= 12, \\
 
a + b + c &= 12, \\
 
ab + ac + bc &= 44, \\
 
ab + ac + bc &= 44, \\
 
abc &= 48.
 
abc &= 48.
 +
\end{align*}
  
 
Then by Vieta's formulas, <math>a</math>, <math>b</math>, and <math>c</math> are the roots of the equation
 
Then by Vieta's formulas, <math>a</math>, <math>b</math>, and <math>c</math> are the roots of the equation
Line 34: Line 31:
  
  
 +
\[
 
\begin{array}{c|c|c|c|c|c}
 
\begin{array}{c|c|c|c|c|c}
 
a & b & c & x & y & z \\ \hline
 
a & b & c & x & y & z \\ \hline
Line 43: Line 41:
 
6 & 4 & 2 & 6 & 2 & 1/2
 
6 & 4 & 2 & 6 & 2 & 1/2
 
\end{array}
 
\end{array}
 +
\]
  
 
Hence, there are <math>\boxed{6}</math> solutions in <math>(x,y,z)</math>. The answer is (E).
 
Hence, there are <math>\boxed{6}</math> solutions in <math>(x,y,z)</math>. The answer is (E).

Revision as of 02:34, 15 June 2020

Problem 30

How many distinct ordered triples $(x,y,z)$ satisfy the equations \[x+2y+4z=12\] \[xy+4yz+2xz=22\] \[xyz=6\]


$\textbf{(A) }\text{none}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }6$

Solution

The first equation suggests the substitution $a = x$, $b = 2y$, and $c = 4z$. Then $x = a$, $y = b/2$, and $z = c/4$. Substituting into the given equations, we get

\begin{align*} a + b + c &= 12, \\ ab + ac + bc &= 44, \\ abc &= 48. \end{align*}

Then by Vieta's formulas, $a$, $b$, and $c$ are the roots of the equation \[x^3 - 12x^2 + 44x - 48 = 0,\] which factors as \[(x - 2)(x - 4)(x - 6) = 0.\] Hence, $a$, $b$, and $c$ are equal to 2, 4, and 6 in some order.

Since our substitution was not symmetric, each possible solution $(a,b,c)$ leads to a different solution $(x,y,z)$, as follows:


\[ \begin{array}{c|c|c|c|c|c} a & b & c & x & y & z \\ \hline 2 & 4 & 6 & 2 & 2 & 3/2 \\ 2 & 6 & 4 & 2 & 3 & 1 \\ 4 & 2 & 6 & 4 & 1 & 3/2 \\ 4 & 6 & 2 & 4 & 3 & 1/2 \\ 6 & 2 & 4 & 6 & 1 & 1 \\ 6 & 4 & 2 & 6 & 2 & 1/2 \end{array} \]

Hence, there are $\boxed{6}$ solutions in $(x,y,z)$. The answer is (E).

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